Problem: Simplify the following expression: $y = \dfrac{-9x^2- 19x+24}{-9x + 8}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-9)}{(24)} &=& -216 \\ {a} + {b} &=& &=& {-19} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-216$ and add them together. Remember, since $-216$ is negative, one of the factors must be negative. The factors that add up to ${-19}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${-27}$ $ \begin{eqnarray} {ab} &=& ({8})({-27}) &=& -216 \\ {a} + {b} &=& {8} + {-27} &=& -19 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-9}x^2 +{8}x) + ({-27}x +{24}) $ Factor out the common factors: $ x(-9x + 8) + 3(-9x + 8)$ Now factor out $(-9x + 8)$ $ (-9x + 8)(x + 3)$ The original expression can therefore be written: $ \dfrac{(-9x + 8)(x + 3)}{-9x + 8}$ We are dividing by $-9x + 8$ , so $-9x + 8 \neq 0$ Therefore, $x \neq \frac{8}{9}$ This leaves us with $x + 3; x \neq \frac{8}{9}$.